Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 5

The Arrhenius equation isHere $E_a$ is the molar activation energy, $R$ the molar gas constant, $T$ the absolute temperature, and $A$ the pre-exponential factor, taken approximately constant over the measured range. For this bimolecular elementary reaction, $A$ and $k_1$ both have units $\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$.

An unweighted least-squares Arrhenius plot using all five data points gives $\boxed{E_a=20.7\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}}$. Define the dimensionless ordinate $y=\ln (k_1/k_{\mathrm{ref}})$ with $k_{\mathrm{ref}}=10^6\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$, and set $x=1/T$. The model isFor increasing temperatures, the five ordinates are $1.82455$, $2.56495$, $3.13549$, $3.52636$, and $4.15888$. The linear regression slope and intercept areConsequently $E_a=-mR=20721\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. The corresponding fitted $A=e^{11.4242}10^6=9.15\times 10^{10}\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. A two-point calculation using just the endpoints gives approximately $21.5\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; the difference reflects the rounded, imperfectly collinear data rather than a unit conversion.

This labelled part introduces the reversible reaction pair without a separate calculation to perform. For the following derivations, define its forward and reverse elementary reaction rates asand the initiating rate as $r_1=k_1[\mathrm{N}{\mathrm{O}}_2][{\mathrm{O}}_3]$. The law of mass action gives $k_2$ bimolecular units and $k_{-2}$ first-order units. Reversibility does not require $r_2=r_{-2}$ while the complete network is reacting: that equality would impose chemical equilibrium on the pair, an extra assumption not needed here.

The quasi-steady-state approximation for the nitrate radical balances its formation against its removal:Equivalently,The net dinitrogen pentoxide production is $r_2-r_{-2}$. Substituting the steady-state balance givesThe reverse rate has been retained throughout and cancels algebraically. The approximation requires sufficiently rapid radical adjustment and fails if, for example, nitrogen dioxide becomes too scarce to sustain the presumed trapping rate. It does not assume that the radical concentration is zero.

Nitrogen dioxide is consumed in reactions 1 and 2 and regenerated in reaction $-2$, soUsing $r_2-r_{-2}\simeq r_1$ from the nitrate-radical quasi-steady-state approximation givesThe factor of two is essential: an initiating event uses one nitrogen dioxide molecule, and the ensuing net radical-trapping event uses a second one. The exact nitrogen balance before making this approximation is $d([\mathrm{N}{\mathrm{O}}_2]+[\mathrm{N}{\mathrm{O}}_3]+2[{\mathrm{N}}_2{\mathrm{O}}_5])/dt=0$.

Adding the initiation step to the net forward trapping step cancels the nitrate-radical reaction intermediate:The net stoichiometry agrees with $-d[\mathrm{N}{\mathrm{O}}_2]/dt\simeq 2d[{\mathrm{N}}_2{\mathrm{O}}_5]/dt$ under the quasi-steady-state approximation.

The remaining fraction is $\boxed{0.120\approx 12.0\%}$. Using the tabulated $300\mathrm{K}$ value rather than replacing it by the fitted value, the constant ozone concentration makes nitrogen dioxide loss a pseudo-first-order reaction:Integrating with initial concentration $[\mathrm{N}{\mathrm{O}}_2{]}_0$ and $t=8(3600)=28800\mathrm{s}$ givesThis uses the nitrate-radical quasi-steady-state approximation over the interval, constant temperature and ozone concentration, and no extra nitrogen dioxide sources or sinks. The stated $\mathrm{c}{\mathrm{m}}^3$ and $\mathrm{mol}\mathrm{c}{\mathrm{m}}^{-3}$ units cancel directly; no factor of $1000$ is needed.